I have yearly precipitation extremes data (data in frequency, 0 to 15) from 1989 to 2019. My hypothesis is that climate extremes (precipitation) increased in last 16 years (2004 to 2019) than previous 15 years (1980 to 2003).
I have performed regression with this data and my results show that the regression line is downward. Then I performed regression using the absolute residuals and this time the results show that regression line of the absolute residuals is upward.
Am I doing right?
Why two kinds of results?
Which one is right?
First of all, why not enter your questions directly instead of putting them into an Excel workbook that we must open?
Second, there is not JMP data table. I could make one from the columns in the Excel workbook, but you do not seem to be using JMP. This Discussion forum is for JMP users.
Here is my answer to your questions.
We seem to be 'going around in circles.'
Thanks for your reply. I use JMP on a regular basis. Sorry, my mistake that I forgot to attach JMP file. I have attached it now that shows that I do use it
The dept. I work with have three licenses/users and they are frequent users.
I am not sure that linear regression is the correct method to compare two time periods. Regression requires matching a X and Y value, like Precipitation versus Year for a trend. What is your response and what is your regressor?
How did you perform both regressions in JMP? Knowing the methods that you used will aid us in helping your with a correct answer.
If you are using counts as a response, then a Poisson loglinear regression might be more satisfactory. But I still don't see how you compared the two time periods. Just by comparing the parameter estimates? If so, there are more comprehensive methods for such a case. Assuming that the trend is linear over time, then you might have a linear predictor like Precipitation = constant + beta(1)*Year + beta(Period) + beta(2)*Period*Year. This way, you can test if there is an average difference between the two periods and if the trend is different between the two periods.
Thanks so much. I have added a excel file which will show you how I have analyzed the data. Your answer will be highly appreciated.
I ran my own permutation test with the data that you supplied. I consistently get a different result. Here is one of them:
So my one-sided p-value would be 1185 / 2501 = 0.4738. That is, it is estimated that 47% of the samples produce a difference at least as large as the observed difference assuming the null hypothesis is no difference (basis for permutations).
Thank you so much for your kind support and advise. The info you provided is very useful. I have done a couple of analyses and looking for some additional advise. Pleased see the attached file. I would appreciate your advsie. Regards
Your residual analysis indicates that the log Y is better suited to the regression because the variance is now constant. Do you know about the built-in Box-Cox transformation? This automated procedure is a generalization of the ad hoc power transformations. The spectrum of parameter values includes the log. A Box-Cox transformation parameter equal to zero is the same as the log. Now when I say the same, this transform does not yield the same value as the log or the power, but they may be interpreted as such. So a parameter value of 2 is interpreted as the square of Y and a value of -1 is interpreted as the reciprocal.
Caution: your wording (thinking) is a bit dangerous. I would not say that there is "no evidence" when the p-value is not below some subjectively chosen threshold (ɑ). I would say that there is insufficient evidence that I would reject the null hypothesis in the test.
Anyway, glad that you are able to progress with your analysis.
Yes, I meant transform the original response. I forgot that you have some count = 0, so this approach won't work. It was just an alternate idea.
I performed the permutation test again a few times. Please see the attached file. My results are totall different than yours, means, I am doing a mistake. I run permutation while I think you did bootstarap. Please advise.
Before I contacted you a Mathematics Prof suggsted me using permutation than bootstrap. Therefore I used permutationt. Also, I do not have an access of Bootstarap.
Also, how you got value of 1185 / 2501 = 0.4738.
My result of 1185 / 2501 = 0.4738 is from 1185 samples out of 2501 total bootstrapped samples exhibited a difference greater than your observed difference.
JMP does not perform the permutation test as far as I know. How did you perform this test with JMP?
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